A particle P is initially at the point with position vector $\cvec{30}{10}$ and moves with a constant speed of 10ms$^{-1}$ in the same direction as $\cvec{-4}{3}.$
(a) Find the position vector of $P$ after $t$ s. [3]
As $P$ starts moving, a particle $Q$ starts to move such that its position vector after $t$ s is given by $\cvec{-80}{90}+t\cvec{5}{12}.$
(b) Write down the speed of $Q.$ [1]
(c) Find the exact distance between $P$ and $Q$ when $t= 10$ , giving your answer in its simplest surd form. [3]
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(a) Let $v=\cvec{-4}{3}k.$ Then speed =$|v|=10=\sqrt{(-4k)^2+(3k)^2}=5k.$ Thus $k=2.$
Hence the velocity=$2\cvec{-4}{3}=\cvec{-8}{6}.$
$\therefore$ position vector of $P$ after $t$ s$=\cvec{30}{10}+\cvec{-8}{6}t$.
(b) Velocity of $Q=\cvec{5}{12}.$
Thus speed of $Q=\sqrt{5^2+12^2}=13$ m/s.
(c) When $ t=10,$ position vector of $P=\cvec{30}{10}+\cvec{-8}{6}\times 10=\cvec{-50}{70}.$
When $t=10,$ position vector of $Q=\cvec{-80}{90}+\cvec{5}{12}\times 10=\cvec{-30}{210}.$
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